Stability of inverse source problem for a transmission wave equation with multiple interfaces of discontinuity

. In this paper, we consider a transmission wave equation in N embedded domains with multiple interfaces of discontinuous coeﬃcients in R 2 . We study the global stability in determining the source term from a one-measurement data of waveﬁeld velocity in a subboundary over a time interval. We prove the stability estimate for this inverse source problem by a combination of the local hyperbolic/elliptic Carleman estimates and the Fourier-Bros-Iagolniter transformation. Our method could be generalized to general dimensions since the weight functions and Carleman estimates are independent of the dimensions.


Introduction
Let Ω k (k = 1, 2, · · · , N ) be a series of bounded connected open sets in R 2 . For 1 ≤ k ≤ N − 1, we assume that (1) Ω k ⊂ Ω k+1 ; (2) The boundary of Ω k (denoted by S k ) is smooth and strictly convex, which means the curvature is positive. We denote the inner (outer) side along S k by S − k (S + k ) and denote the jump of function u across S k by [u] S k , i.e., [u] In this paper, we consider a wave equation with discontinuous principal coefficient where a k are constants satisfying a 1 > a 2 > · · · > a N > 0 and Ω 0 = ∅. More precisely, we consider the following equation where n k denotes the outward unit normal vector of S k . If F ∈ L 2 (Q), then equation (1.1) has a unique weak solution u ∈ C([0, T ]; H 1 0 (Ω N )) ∩ C 1 ([0, T ]; L 2 (Ω N )), see [23]. Here we note that for each F ∈ L 2 (Q), u solves the equation if and only if, for each 1 ≤ k ≤ N , u solves System (1.1)-(1.2) arises naturally in geophysics, more precisely, in seismic prospection of the Earth's inner layers. There is much interest in recovering coefficients or the source term from boundary measurements. Condition a k > a k+1 and the convexity of the interfaces S k make the recovery be possible. In the case when a k < a k+1 or S k is not convex, the rays coming from the inner domain may be trapped and fail to arrive at the exterior boundary due to Snell's law.
In [4], Baudouin et al. prove the Lipschitz stability for a one-measurement inverse problem of identifying the coefficient in the zero-order term for N = 2, by using a global Carleman estimate and the method of Bukhgeim-Klibanov (B-K) [13,14,21]. Later in [25], Riahi proves the Hölder stability for the inverse problem of determining the spatial varied coefficient a(x) in the principle term by two measurements, under an assumption that a(x) is a piecewise constant near the interface. For the one-dimension case, we refer to Bellassoued and Yamamoto [8], in which the authors establish the global Carleman estimate for multiple interfaces. However, the weight function they used cannot work in higher dimension.
The B-K method is powerful for the formulation with a finite number of observations, there are many works on the uniqueness and the conditional stability, and the complete list is too long to be given here. To cite some of them, for example, see Baudouin et al. [3], Bellassoued [5], Bellassoued and Yamamoto [7], Imanuvilov and Yamamoto [18], where the key is to estabilish a global Carleman estimate for each problem. For further details on the Carleman estimate for hyperbolic equations, see for example, see Bellassoued and Yamamoto [8] and Imanuvilov [17]. On the other hand, for Carleman estimates for elliptic or parabolic equations with discontinuous coefficients , we refer to Benabdallah, Gaitan and Le Rousseau [11,12], and Le Rousseau and Lerner [28].
In this article, we are interested in the case when there are more than one interface. Different from [4,25], it is difficult to find a weight function satisfying conditions (a)-(f) listed in Section 2 in the whole domain when N > 2, hence we cannot expect to establish a global Carleman estimate. We have to apply local Carleman estimates which will inevitably create some remainder terms by using cut-off functions. In order to eliminate this remainder terms, we then combine an elliptic Carleman estimate and Fourier-Bros-Iagolniter (FBI) transformation [26,27] which can turn the problem from hyperbolic case into elliptic one. Thus, the main achievements in this paper is that we combine the local Carleman estimates and the FBI transformation rather than the global Carleman estimate to prove the conditional stability of our inverse source problem. We remark that the method in this paper can work for general dimensions since the weight functions and Carleman estimates are independent of the dimensions.
Bellassoued and Yamamoto [6] also use FBI transformation to establish a weak observation estimate which shows the stability in the continuation of solutions from lateral boundary data on an arbitrarily small part of boundary, and prove logarithmic stability in the inverse coefficient problem with single measurement of arbitrarily small part of boundary data, provided that values of coefficients in neighborhood of the boundary are known, see also [9]. By following their strategy, we can determine the source term in our problem by one measurement of the solution on arbitrarily small part of boundary if the source term is known near the boundary.

Notations and the main theorem
Throughout this paper, for a function v = v(x, t) with x = (x 1 , x 2 ) ∈ R 2 , t ∈ (0, T ), we use the following notations: .
For a multi-index α = (α 1 , α 2 ), α 1 , α 2 ∈ N, and X ∈ R 2 , we define where · T denotes the transpose of matrices. For a function g : R → R, we denote g • g(x) = g(g(x)) and g • g • g(x) = g(g(g(x))) and so on. We arbitrarily choose two points x 0 , x 1 ∈ Ω 1 and fix δ > 0 such that Since S N −1 is convex, let y * (x, x j ) be the unique point of S N −1 ∩ {x j + λ(x − x j ) : λ > 0}, (j = 0, 1) and n be the outward unit normal vector of Γ := ∂Ω N . We define the observation region by and set Σ = Γ × (0, T ). In what follows, C > 0 denotes a generic constant which value may change from line to line.
Theorem 1.1. Let u be the solution of (1.1)-(1.2). Suppose that the source term is given by Then if the observation time T is large enough, there exist constants C > 0 and ν 1 , ν 2 > 0 such that the following estimate holds for all γ > 1 .
where g(γ) = e ν2γ , v = ∂ t u is the velocity of wavefield u.
We can assume C is large enough and take in Ω N . The remainder of the paper is organized as follows. In Section 2, we give the local Carleman estimate for both hyperbolic and elliptic cases. In Section 3, we prove the main result, i.e., Theorem 1.1. More particularly, we show the controllability of the source term by the solution and the boundary data in Section 3.1. And in Section 3.2, we show the controllability of the solution by the source term and the boundary data. Then in Section 3.3, we mix them together and prove Theorem 1.1. Section 4 is devoted to proving the Carleman estimate, i.e., Lemma 2.1.

Carleman estimate in an annulus
In this section we present the Carleman estimate for the hyperbolic/elliptic operators in annuluses Ω k \ Ω j (1 ≤ j < k − 1 ≤ N − 1) and Ω k \Ω 0 (2 ≤ k ≤ N ). HereΩ 0 is an arbitrarily fixed neighbourhood of x 0 contained in Ω 1 . For consistency, we use the notation Ω k \ Ω j for 0 ≤ j < k − 1 ≤ N − 1 and when we consider the case j = 0, we are actually concerning about Ω k \Ω 0 . For the following understanding, the (k, j) can be regarded as being fixed.
In order to state the Carleman estimate, we need to choose a suitable weight function ψ(x, t) = d(x) − βt 2 , β > 0 which satisfies some of the following conditions: (d) ψ(x, t 0 ) is a constant along S i at any moment t = t 0 , i.e., ∃c i , s.t.
Note that the condition (e) is valid for the two-dimensional case, but it could be generalized to general dimensions with suitable convexity assumptions [4]. We further set ϕ(x, t) = e λψ(x,t) with λ > 0 and denote Ω k \ (Ω j ∪ S j+1 ∪ · · · ∪ S k−1 ) × (−T, T ) by Q k j . We denote C 3 (Q k j ) the set of functions for which the third derivative is continuous in Q k j . Notice that the weight function ψ(x, t) depends on (k, j). We have the following Carleman estimate with this weight function.
, then there exist constants λ > 0, s 0 > 0 and C > 0 such that for all s > s 0 and u ∈ H 1 Q k j satisfying ∂ 2 t u + a∆u ∈ L 2 (Q k j ) and (2.1), the following estimate holds Remark 2.2. We give the following four remarks on Lemma 2.1.
(1) P 1 w in the hyperbolic case of Lemma 2.1 plays a role in resulting in Corollary 2.3. We will give the proof of Lemma 2.1 in Section 4.
(2) In the case of multiple interfaces, the existence of the weight function satisfying conditions (a)-(f) is still unknown. Or we can say it is hard to construct. However, we can construct a weight function satisfying conditions (a)-(d). In the next Section 3, we will first apply the hyperbolic case of Lemma 2.1 to annulus Ω k+1 \ Ω k−1 (1 ≤ k ≤ N − 1) in Section 3.1, namely an annulus with only one interface. Then in Section 3.2, we will apply the elliptic case of Lemma 2.1 to annulus Ω N \ Ω k−1 (2 ≤ k ≤ N − 1), namely an annulus with multiple interfaces.
(3) In [4], the authors construct a weight function satisfying conditions (a)-(f) when there is only one interface in the annulus. Let us consider annulus We define the weight function where M is chosen to let ψ k be positive. By Proposition 1 in [4] (p. 263), ψ k satisfies conditions (a)-(e), and we choose β = a k+1 δ 1 /3 so that the condition (f) is satisfied.
(4) If we can construct a global weight function satisfying conditions (a)-(f) in the annulus with multiple interfaces, then we can just follow the strategy in [4] to simplify the proof without using the FBI transformation later applied in this paper and improve the stability rate in Theorem 1.1 further.
Proof. By integration by parts we can check that then by the hyperbolic case of Lemma 2.1 we complete the proof.

Stability of the inverse source problem
In this section, we apply Lemma 2.1 to obtain Lipschitz stability of inverse source problem for (1.1). We take the even extension of R and u to the interval (−T, T ). We call this function in the same way and let v = ∂ t u, then v(x, t) satisfies the following system In order to fulfil the condition of Lemma 2.1, we take a cut-off function η(t) ∈ C ∞ (R) which vanishes at t = ±T . We let η(t) satisfy 0 ≤ η(t) ≤ 1 and Recall that So Ω k can be represented by We can further let r 1 satisfy (since Ω 0 is an empty set) Figure 1. The sketch map of the domain D k (r) indicated with pink shaded area.
We define Figure 1 is the sketch map of the domain D k indicated with the pink shaded area. The red circle in Figure 1 is determined by

Control f (x) by y(x, t) and the boundary
First we present the following lemma.
Proof. We take a cut-off function In this annulus, we take the weight function ψ N −1 (x, t) by (2.2) for k = N − 1. Here and henceforth we choose β in the weight function satisfying condition (f) in Section 2. We further choose T N −1 satisfying We set ϕ N −1 = e λψ N −1 and w N −1 = e sϕ N −1 z N −1 . By Corollary 2.3, we know that there exist λ > 0, s 0 > 0 and C > 0 such that for all s > s 0 , we have Without loss of generality, we assume that c 2 Combining this inequality with (3.4), we obtain .

(3.5)
This implies for all s > s 0 , where Now minimizing the right-hand side of (3.6) with respect to s − s 0 > 0, we have where ν N −1 = k1 k1+k2 . The proof of Lemma 3.1 is complete.
In the following we extend the estimation layer by layer and finally cover almost the whole domain Ω N \ B(x 0 , δ). More precisely, we set We will control f (x) in the annulus A k by y(x, t) in O k+1 × (− T 2 , T 2 ). Figure 2 shows the sketch maps of domains A k , O k and O k+1 indicated with pink, brown and yellow shaded areas respectively. Lemma 3.2. For 1 ≤ k ≤ N − 2, there exist constants C > 0, T k > 0 and ν k ∈ (0, 1) such that Proof. Similar to the proof of Lemma 3.1. We take a cut-off function ξ k (x) ∈ C ∞ (R 2 ) satisfying 0 ≤ ξ k ≤ 1 and Then we set z k (x, t) = ξ k (x)y(x, t) and we have and S k (x, t) = −2a∇y · ∇ξ k − ay∆ξ k vanishes outside O k+1 ∪ D k (5r k ) \ D k (6r k ). In order to adapt with the concepts in Section 2, here and henceforth we should replace Ω 0 by D 1 (6r 1 ) when we mention Ω 2 \ Ω 0 .
In this annulus, we take the weight function (2.2) and set ϕ k (x, t) = e λψ k (x,t) . Following the steps in Lemma 3.1, there exists T k > 0, for all T ≥ T k , we deduce Now we combine (3.7)-(3.9) together to get Then by the argument like Lemma 3.1, we can find ν k ∈ (0, 1) such that The proof of Lemma 3.2 is complete.
Proof. For the convenience, we present the sketch maps of domains A k and O k , which are shown in Figures 3a and 3b indicated with gray and pink shaded area respectively. By Lemma 3.1 and Lemma 3.2, let σ = min(ν 1 , ν 2 , . . . , ν N −1 ), we have Then by the Young inequality with p = 1 1−σ and p = 1 σ , we have

Control y(x, t) by f (x) and the boundary
We cannot follow the strategy in Section 3.1 to control y(x, t) due to the extra variable t compared with f (x). In order to make it feasible we need to use the FBI transformation F γ,t0 with respect to the variable t. It is defined by and ω ∈ [−3, 3]. One fact we easily obtain is that, for any x ∈ Ω N , Further noting that for some C 1 > 0. We apply the FBI transformation to (3.1) and get Since H(x, t) is supported in 2T /3 ≤ |t| ≤ 3T /4 and |t 0 | ≤ T 2 + 1, we have for some positive constants C 1 , C 2 . Moreover, we also have there exist constants C > 0 and l k ∈ (0, 1) such that .
Proof. Firstly we take a function φ k (x) ≤ 0 satisfying the following three conditions: In the case when The existence of φ k (x) will be presented in the Remark 3.5 below. Then we definẽ and take a cut-off function χ ∈ C ∞ (R) satisfying 0 ≤ χ(t) ≤ 1 and We set Y (x, ω) = χ(ψ k )F γ,t0 y(x, ω), then from (3.11) we have where [A, B] = AB − BA denotes the commutator. We note that [∂ 2 ω + a∆, χ(ψ k )] is a linear differential operator of the first order and is supported in By (ii) we can verify that 3], then we apply the elliptic case of 3]. There exist constants λ, s 0 > 0 such that for all s > s 0 , we have Therefore, there exist positive constants k 1 , k 2 such that then we minimize the right-hand side with respect to s and complete the proof of Lemma 3.4 with l k = k2 k1+k2 .
Remark 3.5. In this remark, we show the existence of φ(x). We can construct φ k (x) in each annulus Ω j \ Ω j−1 (k ≤ j ≤ N ) according to the following four steps.
(1) When x ∈ Ω k \ Ω k−1 , we set We take a cut-off function 0 ≤ θ(t) ≤ 1 satisfying θ (t) ≥ 0, then in this annulus we define as desired. Moreover, we can verify that so conditions (i) and (ii) are satisfied in this area.
(2) When x ∈ Ω k+1 \ Ω k , we set , we take small enough such that −5 ≤ h 2 (x) ≤ h 3 (x) ≤ −3. Then in this annulus we define We can verify that when (3) When x ∈ Ω j+1 \ Ω j , k < j < N − 1, we define φ k (x) in the same way as (2). We set then we can see that the condition (iii) is also true by the definition of Γ.
Corollary 3.6. For 2 ≤ k ≤ N − 1, there exist constants C > 0 and d k ∈ (0, 1) such that Proof. Let us fix k, for j ≥ k, by Lemma 3.4, we get Hence we can reach the conclusion by using mathematical induction.

C2
, where p k = 1+d k 1−d k and C 1 , C 2 are as in (3.12), such that the following estimate holds true for T ≥ T k : Proof. By (3.10), (3.12) and (3.13) we easily have . (3.14) By Corollary 3.6 and the Young inequality we have for all > 0 We take = e −2C1d k γ , combining with (3.10) and (3.14), then we have .
This completes the proof of Lemma 3.7.
Now in order to finish this subsection we need to show that the H 1 -norm of y in O k × (− T 2 , T 2 ) can be controlled by F γ,t0 y 2 H 1 ( O k ×(−1,1)) . We set y γ (x, t) = F γ,t y(x, 0), then we have where K γ (t) = 2γe −γt 2 .
Proof. By Cauchy Integral Formula, we have for 0 < ρ < 1 and t ∈ [− T 2 , T 2 ], Then integrating this inequality for ρ ∈ ( 1 2 , 1), we get This estimate holds for all x ∈ O k and t ∈ [− T 2 , T 2 ]. Therefore, we have By using the same argument to the first-order derivatives of y γ , we conclude that Then by Lemma 3.7 we complete the proof of Lemma 3.8.
Lemma 3.9. For 2 ≤ k ≤ N − 1 and ∀γ > 0, the following estimate holds Proof. We denote the Fourier transform of y(x, t) for t by y(x, τ ), then by the Parseval identity we have Similarly, we have Therefore, combining with Lemma 3.8 we obtain The proof of Lemma 3.9 is complete.

End of the proof of Theorem 1.1
For simplicity, we set .
As presented in Lemma 2.1, we set ϕ = e λψ , w = e sϕ u.

Then we have
Here (· , ·) L 2 (Q k j ) is the inner product in L 2 (Q k j ). And we write Using the integration by parts as [4] does, we have where I i is the sum of the interior terms And X i in (4.1) is the sum of the remaining interior terms, in such a way that for some constant C independent of s. And B − i in (4.1) is the boundary term on S − i (i.e. ∂Ω i ): and B + i−1 is obtained by replacing S − i and n i in B − i by S + i−1 and n i−1 correspondingly.
We note that there is a minus sign before B + i−1 in (4.1) because n i−1 is the normal vector inward Ω i \ Ω i−1 while n i is the outward normal vector. Hence we have Since u vanishes near ∂Ω j , the last term B + j is zero. The aim is to show that for some positive constant c independent of s.
Then we choose λ large enough such that the term of s 3 λ 4 can absorb the last two terms, and we get (4.2).

The boundary
In both cases ψ satisfies conditions (a)-(d) and [u] Si = a ∂u ∂n i Si = 0.
The proof of Lemma 2.1 is complete.